To solve the equation:

$\tan(3\alpha + 12^\circ) = \cot(\alpha + 36^\circ)$

we can use the identity $\cot(x) = \frac{1}{\tan(x)}$, which gives:

$\tan(3\alpha + 12^\circ) = \frac{1}{\tan(\alpha + 36^\circ)}$

Taking the reciprocal of both sides, we get:

$\tan(3\alpha + 12^\circ) \cdot \tan(\alpha + 36^\circ) = 1$

Now, we have:

$\tan(3\alpha + 12^\circ) \cdot \tan(\alpha + 36^\circ) = 1$

This is a standard trigonometric equation, and we will use it to solve for $\alpha$. Let me work through the next steps to simplify and find the solution.

First, rewrite the tangent product identity: We know that:

$\tan(A) \cdot \tan(B) = 1 \quad \text{implies} \quad A + B = 90^\circ \quad \text{or} \quad n\pi/2$

In our case:

$(3\alpha + 12^\circ) + (\alpha + 36^\circ) = 90^\circ$

Simplifying:

$4\alpha + 48^\circ = 90^\circ$

Subtracting 48° from both sides:

$4\alpha = 42^\circ$

Dividing by 4:

$\alpha = 10.5^\circ$

Thus, the solution is:

$\alpha = 10.5^\circ$

Do you want more details or have any questions?

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Here are 5 related questions:

1. How would the solution change if we had $\cot(3\alpha + 12^\circ)$ instead of $\tan(3\alpha + 12^\circ)$?
2. What happens when $\alpha$ is negative in this kind of equation?
3. How can we express $\tan(x)$ in terms of $\sin(x)$ and $\cos(x)$?
4. What other methods could we use to solve $\tan(A) \cdot \tan(B) = 1$?
5. How would this change in radians instead of degrees?

Tip: In trigonometric equations, converting everything to a single function like $\tan$ or $\sin$ can make the equation easier to solve.